# Simple Question on interpretation of Poisson dist. as approximation to Binomial dist.

I have a question regarding to the notion of using Poisson dist as an approximation to Binomial dist.
I can easily prove $\lim_{n\rightarrow\infty} P(X=k) = P(Y=k)$, where $X \sim$ Binomial(n, p) and $Y\sim$ Poisson($\lambda$).
However, I just couldn't quite get the idea of $\lambda = np$ when $n$ is sufficiently large and $p$ is sufficiently small.
Isn't $\lambda$ is the average number of "successes" over an interval?
How is $np$ the average number of "successes" over an interval?
(Note: $n$ represents the number of identical and independent trials; $p$ is the probability of "success" in one trial.)

Many thanks,
Sebastian

• When you proved your limit assertion, what was $\lambda$? If $p$ is kept fixed, both limits are $0$, which tells us nothing about how the distributions compare. Jun 30 '16 at 20:22
• @AndréNicolas $\lambda$ is the average number of "success" over an interval and what I learnt from the textbook is that we assume $\lambda = np$ , then that limit holds when $n$ are large enough and $p$ are small enough. I believe $\lambda$ is fixed, not $p$. Jun 30 '16 at 20:26
• In a certain time interval, we perform an experiment independently $n$ times, where the probability of success is $\lambda/n$. The number of successes has binomial distribution, parameters $\lambda/n$ and $n$, so with mean $n(\lambda/n)=\lambda$. Jun 30 '16 at 20:30
• @AndréNicolas Oh, I see. Thank you so much. Jun 30 '16 at 20:33
• @AndréNicolas Thanks for the reply but I do know how to prove it Jul 1 '16 at 1:29

## 1 Answer

Here is the setup. In a certain interval, say of time, we perform an experiment independently $n$ times, with probability of success each time equal to $\frac{\lambda}{n}$.

Then the number of successes $X_n$ in that time interval has binomial distribution, with parameters $p_n=\frac{\lambda}{n}$ and $n$. Hence $X_n$ has mean $n\cdot\frac{\lambda}{n}$, that is, $\lambda$.

Your limit calculation presumably showed that $\lim_{n\to\infty}\Pr(X_n=k)=e^{-\lambda}\frac{\lambda^k}{k!}$.